COMEDK 2025 Morning Solved Paper

Concept:In double-slit interference, two waves of equal amplitude $E$ superpose with a phase difference $\theta$.The resultant amplitude is found by vector addition (phasor sum).Explanation:Consider two electric field waves from the slits: each has amplitude $E$.At a point on the screen, they have a phase difference $\theta$.Using phasor addition, the resultant electric field amplitude can be written as:$E_{\text{res}} = E e^{i0} + E e^{i\theta}$.Factor $E$ and use the identity $e^{i0}+e^{i\theta} = 2\cos(\theta/2) \, e^{i\theta/2}$.Thus, $E_{\text{res}} = 2E \cos(\theta/2) \, e^{i\theta/2}$.The magnitude (actual amplitude) is $|E_{\text{res}}| = 2E \, |\cos(\theta/2)|$.Since we consider the amplitude ratio (positive), we take $2E \cos(\theta/2)$.Therefore, the required ratio is $\displaystyle \frac{|E_{\text{res}}|}{E} = 2\cos\left(\frac{\theta}{2}\right)$.Answer:Option D: $2\cos\left(\frac{\theta}{2}\right)$.